Safety Injection during Main Steam Line Break

[mudweez0009]

Why is Safety Injection used during Main Steam Line Break?

All that I can find is that it brings water back to the cold leg to compensate for the loss during the break.

 

[QuantumPion]

Rapid cooldown causes RCS volume to decrease due water density increasing, therefore you need immediate water injection to make up the volume as the volume control system can't handle the change that rapidly. You also risk draining the pressurizer and losing pressure control of the RCS.

 

[mudweez0009]

Thanks that helps a lot! So to clarify.. the steam line break cools the primary system down, which in turn causes primary side to think it needs to generate more power. By doing so, the pressure decreases because it will drain more water from the pressurizer to fill the new steam generator demand.. And so the safety injection system introduces more water so that the pressurizer level can remain stable?

In terms of steam loss, would the same result occur if extraction steam is lost to the feedwater heaters (maybe not necessarily due to a steam line break)? Under normal conditions, the extraction steam is the main source of heat to the feedwater heaters. So I imagine the feedwater heaters would not be able to heat the feedwater enough before it goes back into the steam generator.. But how would colder water in the steam generator affect the power level/pressurizer/etc.?
I'm not so good with the thermo to know if more or less steam would be generated..

 

[QuantumPion]

Thanks that helps a lot! So to clarify.. the steam line break cools the primary system down, which in turn causes primary side to think it needs to generate more power. By doing so, the pressure decreases because it will drain more water from the pressurizer to fill the new steam generator demand.. And so the safety injection system introduces more water so that the pressurizer level can remain stable?

The pressure decreases because the water volume shrinks due to cooling down and increasing density. The reducing water volume is what causes pressurizer level to decrease. Otherwise yes.

In terms of steam loss, would the same result occur if extraction steam is lost to the feedwater heaters (maybe not necessarily due to a steam line break)? Under normal conditions, the extraction steam is the main source of heat to the feedwater heaters. So I imagine the feedwater heaters would not be able to heat the feedwater enough before it goes back into the steam generator.. But how would colder water in the steam generator affect the power level/pressurizer/etc.?
I'm not so good with the thermo to know if more or less steam would be generated..

Main steam line break supposes the steam generator secondary side instantly reducing to atmospheric pressure. Problems occurring with the feedwater heaters are not a big deal and can be handled online by reducing power until the problem is corrected.

 

[jim hardy]

oops a good answer preceded mine...
Well, here 's mine anyway.

.........

Back up a little way in your thinking.

It's all about energy.

What does a steam line break do?
It removes a lot of BTU's from the primary.
That causes the primary to col down, and the water in it to shrink.

What does a primary side break do ? It removes a lot of mass from the primary , and the BTU's carried by that mass.

So the initial symptoms look very much alike - pressurizer level decreases, so do pressure and temperature... and if the steamline break is inside containment pressure there will rise,, so the can two look a LOT alike to the unlucky operator on shift..

For either event the reactor is tripped immediately.
SI will inject boron should pressure get below SI pump pressure, which prevents a return to criticality on cooldown. Return to criticality could put more energy into containment than it can handle, so SI is appropriate for steamline break.
SI provides direct cooling water flow to help cool the core for a primary side break .
So SI is appropriate for both events, secondary as well as primary side break.

..................

Extraction steam is not a really large fraction of total steam flow. Losing it results in cooler feedwater which means you can't boil as much of it with the fixed power available from the reactor. You're making~1200 BTU/lb steam from ~430 BTU/lb feedwater when you have extraction steam, or from around 150 BTU/lb feedwater when you don't have extraction. It's unusual to lose all extraction steam, usually it happens just to one feedwater heater and the effect is only a degree or two at entrance to steam generator..

 

[mudweez0009]

Edit: This now goes for either of you.

Well for the second part, let's say it is NOT because of a steam line break (So the scenario is, for some unknown reason, there is no extraction heat from the turbine for the feedwater heaters). I guess what would be a direct result in terms of the pressure/power/etc., assuming that it isn't just handled online? aka, the turbine governor valves don't reposition, or anything else, what happens to pressure/power?

I figure that the amount of feedwater turning to steam in the steam generator (S/G) would decrease if the feedwater is colder (because of the now bigger delta T between the feedwater and the coolant from the hot leg). Since less steam is created in the S/G, less steam goes to the turbines. If less steam is going to the turbine, and the turbine governor valve doesn't reposition, would the power increase or decrease?
If the steam demand stays the same, then I imagine the power would increase to generate more steam to meet it..?

Btw, my reason for asking: I'm currently taking a nuclear engineering course which intends to teach us the plant integration. I have NO background in nuclear or mechanical. I am trying to get a grasp of basic concepts so I can speak at least somewhat coherently on the topics, so I have a decent amount of questions. Would you be at all inclined to assist in others? Logically talking through these helps a lot.

 

[jim hardy]

Cooldown of feedwater is an event covered in the accident analysis chapter of pwr FSAR's.Initially it would cool down the reactor cold leg, which would cause the reactor power to increase.
It would also shrink steam generator level causing feedwater regulating valves to open, making things worse. .
That'd be the natural tendency of the plant.

So the instrument system has something called "overpower runback" which senses reactor power as difference between hot and cold leg temperatures. Temperature rise is proportional to power.

When cold leg starts to plummet it looks like a reactor power increase is imminent, so an "overpower runback" is initiated.
That runback drives the turbine throttle valves in closed direction, reducing steam demand to match what the reactor can deliver. Ours would pulse the throttle valves until temperature rise across reactor returned to normal.

Once again, it's a balancing act with energy.
Keep energy in back of your mind when visualizing these systems - the plant is just moving energy from the fuel to the grid via first BTU's in hot water, next steam, next mechanical torqueXrpm, and finally electrical kilowatts.
That''s why a plant requires just a handful of Nuclear engineers but scores of mechanicals and electricals. The reactor runs fine for it was developed by geniuses in WW2. The rest of the apparatus that moves the energy out to real world
dates back to late 1800's. Even Titanic had a steam turbine.

if you don't mind my unprofessional qualitative talk-throughs, I'm honored to help. Mentors please advise if I'm not up to PF standards here. My math has got rusty and my Latex was never good - but think i can help beginners with these basic concepts.

old jim

ps - see also http://www.westinghousenuclear.com/Products_&_Services/docs/flysheets/NS-ES-0030.pdf

old jim

 

[QuantumPion]

Edit: This now goes for either of you.

Well for the second part, let's say it is NOT because of a steam line break (So the scenario is, for some unknown reason, there is no extraction heat from the turbine for the feedwater heaters). I guess what would be a direct result in terms of the pressure/power/etc., assuming that it isn't just handled online? aka, the turbine governor valves don't reposition, or anything else, what happens to pressure/power?

Feedwater temperature would decrease, thus slightly increasing the reactor power. The turbine power would remain the same. Thus the net effect is lower plant efficiency (which is the purpose of feedwater heaters).

I figure that the amount of feedwater turning to steam in the steam generator (S/G) would decrease if the feedwater is colder (because of the now bigger delta T between the feedwater and the coolant from the hot leg). Since less steam is created in the S/G, less steam goes to the turbines. If less steam is going to the turbine, and the turbine governor valve doesn't reposition, would the power increase or decrease?
If the steam demand stays the same, then I imagine the power would increase to generate more steam to meet it..?

In a PWR, the reactor is a slave to the turbine. While the turbine governor valves remain constant, the steam demand will be constant. If you reduce the temperature of the feedwater, it will extract more heat from the primary side in order to keep the same steam pressure. Removing more heat from the primary side cools it down, which thus increases reactor power due to the negative moderator temperature coefficient.

 

[mudweez0009]

Okay so my answer was correct, but for not exactly a correct reason. Thank you both!
I just have a lot of questions dealing with the steam/turbine/feedwater portion of the plant (as I feel like its the main "big picture" concept to grasp). For example, things we discussed are generator tube ruptures, main steam header (which I can find hardly anything online about), and :

1.) What happens during a steam generator tube rupture? How would this get resolved (either by the operator or the plant systems itself)?

My knowledge: A S/G tube rupture means the tubes inside the S/G which contain the radioactive coolant. If this gets mixed in with the steam in the generator its a huge problem. Therefore the first step would be to cut off the steam to the turbine by closing the Main Steam Iso Valves (MSIVs). No steam is going to the turbine and the reactor is shut down. What I don't know is what steps would be taken to startup again, or resolve it without shutting down the plant..

2.) How does main steam header pressure affect reactivity?

My knowledge: Not too sure what a steam header even does other than serve as a middle man between the S/G's and the turbine, and between the S/G's and the condenser. I would think that if the header experiences higher pressure, that means more steam is coming from the S/G than the turbine needs. Therefore, more steam would be sent to the condenser. Not sure how reactivity would be affected in this case.
Same but opposite goes if the main header pressure is low. I would believe that the there isn't sufficient steam supply to the header based on turbine demand. I However I think that this would cause the reactor to generate more steam, which results in an increase in reactivity.

 

[QuantumPion]

Okay so my answer was correct, but for not exactly a correct reason. Thank you both!
I just have a lot of questions dealing with the steam/turbine/feedwater portion of the plant (as I feel like its the main "big picture" concept to grasp). For example, things we discussed are generator tube ruptures, main steam header (which I can find hardly anything online about), and :

1.) What happens during a steam generator tube rupture? How would this get resolved (either by the operator or the plant systems itself)?

My knowledge: A S/G tube rupture means the tubes inside the S/G which contain the radioactive coolant. If this gets mixed in with the steam in the generator its a huge problem. Therefore the first step would be to cut off the steam to the turbine by closing the Main Steam Iso Valves (MSIVs). No steam is going to the turbine and the reactor is shut down. What I don't know is what steps would be taken to startup again, or resolve it without shutting down the plant..

The defective tubes are identified using eddy current testing, and are then plugged. There is a limit as to how many tubes may be plugged without significantly affecting the total heat transfer. If too many need to be plugged, the plant must reduce its maximum power or replace the tube sheet (expensive).

2.) How does main steam header pressure affect reactivity?

My knowledge: Not too sure what a steam header even does other than serve as a middle man between the S/G's and the turbine, and between the S/G's and the condenser. I would think that if the header experiences higher pressure, that means more steam is coming from the S/G than the turbine needs. Therefore, more steam would be sent to the condenser. Not sure how reactivity would be affected in this case.
Same but opposite goes if the main header pressure is low. I would believe that the there isn't sufficient steam supply to the header based on turbine demand. I However I think that this would cause the reactor to generate more steam, which results in an increase in reactivity.

The main steam header is just the point where the main steam lines for each steam generator meet and combine. Remember, the reactor is slave to the turbine. Changes to steam pressure will affect the steam generator heat transfer rate and thus reactor power. The turbine gets exactly the steam it demands (unless the steam generators dry out - bad). Increasing turbine power lowers steam pressure, which increases steam generator boiling until pressure reaches new equilibrium. If turbine power is lowered, steam pressure increases, reducing boiling until new equilibrium is reached.

 

[mudweez0009]

For your response to #1, could you elaborate more? I guess in PWR's there are more than 1 steam generator, so they could shut that S/G down for maintenance, but I imagine there are a large number of steps that have to be taken if something like this happens, such as certain valves are shut (I mentioned the MSIV above), different processes are stopped (like the turbine), different processes are started, etc.

For your response to #2, I guess it wasnt clear enough to make me sure that I'm correct in my logic. Can you work through it but using a BWR instead (since in BWRs, turbine is slave to the reactor)?

If you haven't noticed, I'm really trying to UNDERSTAND the flow of processes and their consequences by stepping everything out. Like for #1, in a PWR, if a generator tube ruptures, the operator has to do certain tasks, and those tasks will have an effect on other things. For #2, the main steam header in a BWR is experiencing high (or low) pressure, this means the S/G has to react a certain way, which affects reactivity a certain way.

 

[jim hardy]

This link has links to the NRC training materials for the introductory course they give to their inspectors, i think.

I tried to download them but am on a weak server here and it crashed .

http://www.law.cornell.edu/wayne/projects/entergy/callcenter_website/orient_train/ont_nuclear_power_for_energy.htm

aha this one loaded, it's a pretty good bird's eye introduction.
http://www.nrc.gov/reading-rm/basic-ref/teachers/04.pdf

QPion's answers are good, and i admire his direct answers. Makes me feel my thinking is muddied by age. Maybe too many late nights in the plant...

Anyhow - big picture on SG tube leaks is this -

The leaks usually start out small. Your first indication is count rate at the condenser air ejector starts to climb, and this is long before radioactive nuclides become noticeable in secondary water samples .
So you start getting ready for a shutdown.
When water samples can tell you which generator is leaking you'll shut down reactor and begin a cooldown, reducing reactor pressure as soon as practical to minimize leakage.
In old days we could isolate affected generator's secondary and pressurize it during cooldown so leakage goes other way - clean water into reactor instead of contaminated water into secondary systems. I don't know if that's done anymore.

Once plant is cooled down you drain primary and locate the leaking tube(s), plug both ends , and probably do a lot of testing to see if any other tubes are thinned or near failure.

It's a big operation, a week minimum and often a month.

I'm not aware of any leaks that were ever big enough to affect reactivity. One gpm would be a big one where i came from.

That training material should help you with mental pictures of the parts and terminology.

Ask your questions, but be aware it'd take a book to describe the systems and that book is already written.

PWR Steam Header Pressure affects reactivity only through temperature, look at saturation curve for water substance (In my days Civils took Thermo, and Electricals took Fluid Flow - i think so that we'd be still able to communicate after graduation.) It's back to that energy flow - heat transfer through tubes is quite good, pressure variation of secondary translates directly to temperature (saturated liquid), and delta-temperature across tubes defines heat flow out of primary...
both sides obey the properties of water in the steam tables. Secondary is saturated, primary is well sub-cooled.
As we said in the control room: " It's only water. Keep it that way."

In a BWR header pressure affects reactivity via the tiny steam bubbles in the boiling water - more steam bubbles shut down the reaction, fewer steam bubbles enhance it. That's why the control system for a BWR is pressure based vs PWR's temperature based.

If you haven't had Thermo, suggest to professor he spend one day on navigating the steam tables .

Enjoy your course. You know, nuke plants need civil engineers too... If you enjoy this course, interview your local utility...

happy to address your questions. Maybe Qpion and i can complement one another's answers.old jim

 

[Hiddencamper]

For your response to #1, could you elaborate more? I guess in PWR's there are more than 1 steam generator, so they could shut that S/G down for maintenance, but I imagine there are a large number of steps that have to be taken if something like this happens, such as certain valves are shut (I mentioned the MSIV above), different processes are stopped (like the turbine), different processes are started, etc.

For your response to #2, I guess it wasnt clear enough to make me sure that I'm correct in my logic. Can you work through it but using a BWR instead (since in BWRs, turbine is slave to the reactor)?

If you haven't noticed, I'm really trying to UNDERSTAND the flow of processes and their consequences by stepping everything out. Like for #1, in a PWR, if a generator tube ruptures, the operator has to do certain tasks, and those tasks will have an effect on other things. For #2, the main steam header in a BWR is experiencing high (or low) pressure, this means the S/G has to react a certain way, which affects reactivity a certain way.

In a BWR, the turbine is slaved to the reactor, as you said. BWRs are analyzed such that they must be above about 750 PSIG in order to exceed 25% power. They are not analyzed to be producing power at lower pressures. Additionally, BWRs have a main steamline break protection, such that if the reactor senses steamline pressure low, the MSIVs will automatically fast close, generating a reactor scram if the reactor mode switch is in RUN (and even if you are in STARTUP mode, you'll likely trip out on high flux). In general, BWRs are operated between 970-1050 PSIG, depending on the specific plant, the specific core design, etc.

That said, BWRs have positive pressure feedback when they are steaming (which is almost all the time, except during cold startup). BWR operations in the US do not allow direct reactivity control using steam header pressure. I've done it before in my BWR's simulator, and you do see a response, but the moderation of pressure to affect reactivity is explicitly not allowed as it is not an analyzed reactivity control method for BWRs and can lead to reactivity transients. (There may be a plant out there that does have permission to raise pressure at end of cycle conditions to help boost reactivity, but it is not the norm).

Generally, pressure in a BWR is set constant. At my plant, we hold about 920 PSIG at the main steam equalizing header. This equates to about 1020 PSIG in the reactor. As power is raised or lowered, the change in steamflow causes the pressure in the main steam equalizing header to raise or lower. The turbine will respond by automatically opening or closing control valves to maintain pressure in the header at its set point, which in turn, maintains reactor pressure around its setpoint, and ensures that pressure is not controlling reactivity.

During heatup, you do see effects on reactivity. If you do not have adequate steam removal, you'll have pressure come up, power come up, pressure come up faster, power come up faster, and you'll either swell your water inventory causing a high level trip or steam line closure, trip out on high flux, or even worse, violate your tech spec heatup rate.

When you are at power, any sudden change in throttle rate (how far open or closed your control valves are) will have large effects on both reactivity, actual reactor water level, and indicated reactor water level. For example, if a bypass or control valve were to suddenly jump full open, you will have an increase in steam flow and a decrease in pressure. In the reactor, your water level indications will rapidly rise, and power will decrease. The other control and bypass valves will automatically reposition to maintain steam header pressure, which halts the transient. If they didnt reposition, the water level swell would continue and would trip your main turbine which would then result in a reactor trip. If you had a bypass or control valve suddenly go full closed, you would see water level decrease, power increase, and again your control/bypass valves would respond to the transient. If they did not respond, the reactor pressure would increase and you would trip out on high pressure. If the high pressure scram failed, you would trip out on high flux.

This brings another big point, any time you lift safety valves, or perform any non-smooth pressure control action, you will have sudden swings in actual and indicated level. This is very challenging to deal with, as you tend to hit your high and low water level trips and you are dealing with major equipment going offline as a consequence. It takes a bit of training to get used to to these responses.

 

[mudweez0009]

First, thanks so much for the response, it is really informative.

For example, if a bypass or control valve were to suddenly jump full open, you will have an increase in steam flow and a decrease in pressure. In the reactor, your water level indications will rapidly rise, and power will decrease.

I have 2 questions about this part you said..

1.) I'm confused on how an increase in water level causes the power to decrease? Wouldnt the BWR reactor "want" to produce more steam if the water levels were high, in which case would increase power? (lets assume the turbine is malfunctioning which is causing the pressure fluctuations).

2.) The relationship between power and reactivity is that if power goes up, reactivity goes up...and power goes down, reactivity goes down. And neither one is especially bad if it means trying to get it back to criticality. Correct?

 

[Hiddencamper]

First, thanks so much for the response, it is really informative.



I have 2 questions about this part you said..

1.) I'm confused on how an increase in water level causes the power to decrease? Wouldnt the BWR reactor "want" to produce more steam if the water levels were high, in which case would increase power?

2.) The relationship between power and reactivity is that if power goes up, reactivity goes up...and power goes down, reactivity goes down. And neither one is especially bad if it means trying to get it back to criticality. Correct?

For 1:

If you were to add water to a reactor, you get a slight improvement in subcooling. This results in a slight increase in power, (but also results in a net decrease in plant efficiency). If you were to decrease water in a reactor, you get a slight decrease in power, until you uncover the feedwater spargers, at which point you get a large decrease in power as the incoming water is not adequately subcooled.

What I was talking about though. Is when you have a pressure change, the sudden pressure change will result in a change in saturation pressure/temperature. The overall effect is a decrease in pressure will cause MORE voiding, which causes your water density to DECREASE. A decrease in density results in an increase in level. Additionally the pressure change will affect your reactor water level sensing lines causing indicated level to doubly increase. On a pressure increase the opposite happens. In this case, the water level change is due to a change in density. Power is affected due to void content change.

For #2:

Reactivity is like "Acceleration" in traditional kinematics. A critical reactor has 0 reactivity. If you have positive reactivity, you are super critical and neutron flux is increasing. Negative reactivity, flux is decreasing. The rate of change is exponential. Eventually, as flux increases, power increases, you produce more heat, and you boil more water (generate more voids). The increase in void content reduces moderation, which removes reactivity until it is back to 0. If you add negative reactivity, power will decrease exponentially, and the reactor is subcritical. The reduction in flux then reduces power, reduces heat output, and ultimately reduces void content. This improves moderator density, brings reactivity back to 0, and the reactor goes to steady state at a lower power level. The exception to this is if you add a LOT of negative reactivity at once (like in a SCRAM), the reduction in void content and other effects in the core won't be able to make up for the large reactivity reduction, leaving to the reactor shutting down. If you had a large reactivity increase, the scram system will detect high flux and automatically actuate to stop the increase, bringing the core subcritical.

Hope this helps.

 

[mudweez0009]

Okay that makes sense, thank you! Out of curiosity, what is your experience? Operator? Professor? I may just have to contact you again in the future with more questions as they pop up, since you seem to have a wealth of knowledge. I like the whole "if this happens, then this happens" step-through approach, as it really helps understand the concepts and theory that general manuals and literature online seem to lack.

 

[Hiddencamper]

Okay that makes sense, thank you! Out of curiosity, what is your experience? Operator? Professor? I may just have to contact you again in the future with more questions as they pop up, since you seem to have a wealth of knowledge. I like the whole "if this happens, then this happens" step-through approach, as it really helps understand the concepts and theory that general manuals and literature online seem to lack.

Nuclear engineer for several years. Currently getting a senior reactor operator license at a BWR. Feel free to ask questions. These are the fun questions to answer.

 

[QuantumPion]

2.) The relationship between power and reactivity is that if power goes up, reactivity goes up...and power goes down, reactivity goes down. And neither one is especially bad if it means trying to get it back to criticality. Correct?

If power goes up, then reactivity goes down. But you need to insert positive reactivity to make power go up initially. If reactivity went up when power went up, you'd have uncontrollable positive feedback.

 

[Hiddencamper]

If power goes up, then reactivity goes down. But you need to insert positive reactivity to make power go up initially. If reactivity went up when power went up, you'd have uncontrollable positive feedback.

BWRs have a positive pressure coefficient. Normally, the turbine and/or steam bypass valves will automatically modulate to control pressure. If pressure control is lost, you will have a pressure change which will affect flux, and you will have a run-away effect, where pressure causes power to go up, which causes pressure to go up.

The worst case scenario for a typical BWR is a fast closure of all main steam line isolation valves. The pressure spike will result in a prompt peak of around 210% flux, which is arrested by the lifting of safety valves and the reactor scram. If the scram fails, the lifting of safety valves will reduce pressure and your ATWS (Anticipated Transient without scram) logic will trip your recirculation pumps to drop power down to about 40%.

In the event pressure control is lost, operators are supposed to manually scram the reactor. It's not a requirement, but you will likely end up either in an unanalyzed condition or, more likely, with an automatic reactor scram.

 

[mudweez0009]

That's cool Hiddencamper, and good luck on the license.
Thank you Quantum.

Okay, I imagine this question is a simple thermo one, I just don't know thermo. In a heat exchanger, what would happen to the heat transfer rate if there are particulates/solids in the tube side of the heat exchanger?

This is part of a much bigger picture I'm trying to paint, which involves the main condenser and particulates in the tube side, the resulting efficiency, etc. But I'd probably figure that out on my own if I knew whether the steam would condense less or more because of particulates in the tube side.

I imagine the heat transfer rate decreases, because the particulates take up space that the cooler water would normally occupy. That means the heat exchanger won't function as properly (less steam condensed => hotwell temps increase => eliminates vacuum => plant efficiency decreases)..

 

[Hiddencamper]

You're right.

That's also why my plant will inject chlorine and a cocktail of other chemicals into the circulating water passing through the condenser. To reduce scale buildup, biological fouling, and corrosion of tubes. Just before going into summer months, where heat transfer is most crucial to maximizing thermal efficiency, we will do chemical treatments of the condenser tubes to "shock" them, and try to break up as much of the stuff as we can.

The resulting loss of heat transfer will increase the pressure in your condenser, and increase the temperature of your condensate in the condenser hotwell (where the condensed liquid goes). The increased pressure in the condenser reduces the amount of work which can be extracted by the main turbine. The increased temperature of the hotwell fluid could impact your NPSH (Net positive suction head) on your condensate pumps. The increased temperatures can also affect your air ejector/deairation systems. In all cases, if you start to bump into a limit you'll need to reduce plant thermal power to recover.

 

[mudweez0009]

Here's a question:
How do fission product neutron poisons, such as Xe, affect the reactor(?) Tave? We have been discussing Thot (Th), Tcold (Tc), and Tave, and how the operators shift Tave in order to fix Th at a thermal limit. We were shown a graph that had Tave increasing with increasing power, because a higher Tsteam is desired. I have 2 possible scenarios below.

Scenario 1.) When power is increased for general purposes (high demand), Xe concentration will initially decrease. Xe concentration will reach a minimum until it starts to increase to a new, higher equilibrium based on the higher power. So Tave will initially increase with the initial decrease in Xe. Then the Tave will start to decrease as Xe starts to increase to its new equilibrium.

Scenario 2.) When power is decreased for general purposes (maintenance), Xe concentration will initially increase. Xe concentration will reach a maximum until it starts to decrease to a new, lower equilibrium value based on the lower power. So Tave will initially decrease with the initial increase in Xe. Then the Tave will start to increase as Xe starts to decrease to its new equilibrium.

For both scenarios (depending on if I'm correct on the Tave part), what does the operator have to do to control these effects?

 

[Hiddencamper]

You do need to adjust for xenon. I'm not as familiar with PWRs, I believe Tave is typically controlled by adjusting boron, while deltaT is adjusted by load.

For a BWR, after a small load drop or increase, over the next 12 hours you'll see several "raised core flow" "lowered core flow" entries in the operating logs, in very small increment, to make up for xenon changes. For larger load drops or increases, you'll probably need to notch rods as well. For example, after a large power increase, the initial decrease in xenon will raise your rod line. (Rod line is a measure of how much power you have at 100% core flow, and has a limit). Xenon changes affect rod line, and if rod line starts increasing you'll need to notch insert a little bit to push it back down.

 

[mudweez0009]

Hmmm.. We primarily focus on PWRs in this class so I would prefer the understanding based on those. However, would the processes in a BWR be that much different for these specific scenarios?

 

[jim hardy]

Have you done a "four factor" or "six factor" reactivity calc yet ?

Xenon just decreases the probability that a neutron will make it to an absorption-fission reaction.
So it's negative reactivity.

PWR's are basically all rods out operation these days, so we'd have to dilute the boron a little to add equal amount of positive reactivity. It's only a problem near end of core life where there's not a lot of excess reactivity left in the fuel.

As you'll doubtless study, xenon gets burnt out by the high neutron density at full power. It's only after shutdown(or a large power reduction) when the iodine fission fragments decay into xenon that it impairs startup.

It's a simple differential equation rate problem that you'll probably study in your class. Iodine decays into xenon at one rate, and xenon decays away at another rate, so after shutdown there's a time-to-peak xenon that's a matter of hours. If you want to restart it may be necessary to do so either before or after peak.

We operate on a programmed T-Average, designed to keep Tcold nearly constant.
That's a compromise - the turbine would like to get constant temperature steam, which would require T-cold to rise with power.
The reactor would like to see constant T-average, which would require T-cold to drop and T-hot to rise with power;
so the designers decided on sliding T-average. It let's them have Thot around 600 degrees which allows reasonable operating pressure.

When you have a lot of xenon you may not be able to reach full power because the negative reactivities from increasing T-ave and Xenon can add up to more than what's left in the fuel.
I can't recall being unable to achieve criticality, but i can remember waiting on xenon to decay so as to not generate so much wastewater by a big dilution.




ols jim

 

[mudweez0009]

Have you done a "four factor" or "six factor" reactivity calc yet ?

We have not done those calcs, and we wont. This class is all about THEORY of plant integration.

Xenon just decreases the probability that a neutron will make it to an absorption-fission reaction.
So it's negative reactivity.

PWR's are basically all rods out operation these days, so we'd have to dilute the boron a little to add equal amount of positive reactivity. It's only a problem near end of core life where there's not a lot of excess reactivity left in the fuel

So let me get this straight. In terms of a power reduction, the Xe initially increases to a maximum amount. During this time, Tavg increases, and the boron concentration is diluted in the reactor makeup system in order to add positive reactivity back into the reactor to balance out the negative reactivity caused by the Xe increase.
Then, as the Xe decreases back to equilibrium, Tavg decreases, the operator must increase boron concentration.

Is there anything else that happens? You mentioned PWRs like to have control rods fully inserted, so withdrawing/inserting them is not an option, but is there a 3rd, 4th, etc. option?

 

[jim hardy]

We have not done those calcs, and we wont. This class is all about THEORY of plant integration.




So let me get this straight. In terms of a power reduction, the Xe initially increases to a maximum amount. During this time, Tavg increases, and the boron concentration is diluted in the reactor makeup system in order to add positive reactivity back into the reactor to balance out the negative reactivity caused by the Xe increase.
Then, as the Xe decreases back to equilibrium, Tavg decreases, the operator must increase boron concentration.

Is there anything else that happens? You mentioned PWRs like to have control rods fully inserted, so withdrawing/inserting them is not an option, but is there a 3rd, 4th, etc. option?

hmmm if you're not studying criticality and math of multiplying systems, i think prof is missing something...

First , a clarification::: T average is maintained by matching reactor power against electrical power. Xenon does not control Tave. We control power into the turbine with its inlet valves, and we control reactor power with either rods(during power ascension) or boron (steady state).

So let me get this straight. In terms of a power reduction, the Xe initially increases to a maximum amount.

During power operation you build up iodine which is a precursor to xenon. Iodine doesn't affect reactivity much, but it is there. That iodine decays to xenon, which gets burnt out by the high level of neutrons at power. So your equlibrium xenon isn't very high.
Iodine too has an equilibrium value, but since iodine isn't a great neutron poison it's only significance is as the xenon precursor.

At equilibrium conditions, rate of xenon production(some directly from fission, some by decay from iodine) = rate of xenon removal (from burnout by neutrons and by decay)

Now upon a significant reduction in power, the rate at which xenon gets burnt out by neutrons decreases. But you have a substantial equilibrium iodine that is still decaying into xenon, so the rate of xenon production is now greater than the rate of xenon removal. So xenon first builds, then diminishes - it's a rate problem just like those tank problems you solved in differential equations course.

If xenon gets so high that the reactor can no longer maintain power, we'd have to reduce electrical load to match whatever the reactor can deliver. That's happened to us.
But Tave is maintained by that matching of reactor and turbine powers mentioned earlier.

Tave is defined by how much steam is arriving at the turbine.
We measure steamflow through the turbine (from its first stage pressure).
Steamflow sigal undergoes a y=mX+B function to get transformed from lbs/hour of steam to degreesF of desired Tave.
My particular plant was Tave = 547 at no load(zero steamflow) and 574 at full load as [STRIKE]measured at[/STRIKE] defined by the turbine first stage pressure(which is proportional to steamflow).. Full load was around 9.6 million pounds per hour of steamflow. For perspective - a residential swimming pool might be 20,000 gallons and we boiled that much water into steam every thirty seconds. This is big machinery.

Temperature has a reactivity worth, so we can swap temperature reactivity for xenon reactivity, reduce power so Tave will be lower(still match reactor and turbine though) and avoid diluting. Or we could dilute... either gives a measure of positive reactivity whic will offset the xenon's negative reactivity.


As xenon goes away through decay or burnout or both, reactivity comes back. We could let power come on up, Tave will follow, again swapping temperature for xenon. Or we could borate. Or , at reduced power we could use rods so as to not make so much wastewater that has to be handled.

Sounds confusing i know, but it's quite practical and becomes reflexive.

You mentioned PWRs like to have control rods fully inserted, so withdrawing/inserting them is not an option, but is there a 3rd, 4th, etc. option?

That's withdrawn, not inserted. PWR rods go in from the top not the bottom as on BWR's.
Full out assures that the neutron flux distribution is natural and smooth from bottom to top of core. Inserting rods squishes it down toward bottom, with result that bottom of core works harder than the top. Reactor engineers like an even distribution ( they call it F(z),,) so we operate most of the time with rods all the way out.

We have not done those calcs, and we wont.

Now it's unfortunate that you aren't studying a little reactor physics.

TRy this simple concept for heat production in the reactor:
Swap your thinking of the process from continuous to stepwise, like the individual frames in a motion picture.
In any given frame there are X neutrons present.
At steady state, in the next frame there will also be X neutrons present.
So between frames, every neutron got captured, produced 2.2 more; and of those 2.2neutrons, 1.2 of them leaked away or got absorbed by something like u238 or structural steel or water or xenon or whatever, leaving X neutrons every generation.
THAT is exactly "critical" - the ratio of neutrons in succeeding generations is 1 to 1.
It's a multiplying system, and at critical you are multiplying by 1, AND clearly 1X1x1xx1 = 1 so everything is steady state.
That multiplying factor is called K_effective . It is of great import to reactor engineer types.
It is always very near one. Greater than 1 is "supercritical", less is "subcritical".
The four factor formula i mentioned calculates Keff as a product of probabilities ,
more here http://en.wikipedia.org/wiki/Four_factor_formula

so when you hear about "reactivity" they are talking about Keff. Positive reactivity increases Keff, negative decreases it.

For the reactor, the time steps we would use are in milliseconds to microseconds range and Keff hovers quite close to 1.0. At 1.007 the reactor runs away from you as happened at Chernobyl. There are plenty of built in features to avoid even approaching that possibility.

I feel like I've rambled. Run this by professor - maybe he'll see from your questions that his course needs a vocabulary preface.

Hang in there - your inquisitiveness belies intelligence and interest. I just loved working around the machinery, and the people in nuclear power are exceptional. See if you can get a course in nuclear engineering - if you passed vector calculus you can pass reactor physics. You're a civil, right? Buckling is buckling be it in a column or in neutron flux, exact same equations.

old jim

ps there are a lot of introductory courses on the 'net. here's one (of many), i haven't culled them.
ftp://ftp.ecn.purdue.edu/jere/BURN/Ch-04.pdf

 

[jim hardy]

Here's another thread you might peruse. It has a little more detail on multiplying systems.

https://www.physicsforums.com/showthread.php?t=513412Poster "Morbius" is a genuine professional nuclear physicist, while I'm just an old instrument guy.

 

[mudweez0009]

Wow Jim, thanks so much. You, Hidden, and Quantum really have been doing a tremendous job helping me to understand this stuff. I'm not a person that likes just an answer, because I HATE not knowing something, and it will be the death of me for sure! Even in my undergrad, I would get the solutions from the TA and sit there for hours until I understood the problems, not just take 10 minutes to copy. So your responses, although you feel are rambling, are simple to understand, and that's what matters. There are tons of textbooks out there that just list equations and proofs, but there are hardly any that explain things well.

That's withdrawn, not inserted. PWR rods go in from the top not the bottom as on BWR's.

Yes, I don't know why I said "inserted". Long day.

My hope for using this forum was to get people with experience that actually want to help, and I definitely need to keep utilizing you guys throughout this course and in the future. I'm actually getting my masters in mechanical engineering, I just wanted to try a nuclear course this semester to see if its a field I'd like to pursue. It has probably been the most interesting, but it also is very challenging.

Thanks for the links as well, I've been Googling non-stop throughout this course to find information, so anything else will help.

 

[jim hardy]

glad to help. makes an old 'secondhand lion' feel less useless. Thanks for the kind words.

good luck in your studies and career. Nuclear power was good employment for my working years and i believe it will be necessary to get us over the energy hump that faces your generation.

If you find this interesting, check into what your school has available.
Best thing i ever did was take a course called "reactor operation" where we started up , maneuvered, and shut down the school's little swimming pool reactor. It was about the size of a washing machine, under thirty feet of water.

http://reactor.mst.edu/

Learned its simple systems and mapped the flux distribution in the core. Gave me a running start for career as instrument guy in a nuke plant.

There is a certain psychological makeup that makes for a good operator , and that is the ability to remain detached and keep attuned to big picture . I am a detail oriented guy who gets obsessively focused , and would not make a good operator. No shame in that though - i was sought after as a troubleshooter for those weird problems that throw out confusing symptoms. Point being - find a niche where you can contribute meaningfully and become really good at whatever you do. You will be rewarded.
I use a steamship analogy - Everybody doesn't work on the bridge in a white uniform. Titanic's coal shovelers got good spots on the lifeboats as rowers because of their strong arms . Be good at whatever you do, no matter how paltry it seems at first.

old jim

 

[mudweez0009]

Hey! So were going back to a steam line break scenario again..

Why is a steam line break more serious at end of core life than at beginning? I assume its based on the Moderator temp coefficient (MTC) which also changes from beginning of life (BoL) to end of life (EoL)..

 

[QuantumPion]

Hey! So were going back to a steam line break scenario again..

Why is a steam line break more serious at end of core life than at beginning? I assume its based on the Moderator temp coefficient (MTC) which also changes from beginning of life (BoL) to end of life (EoL)..

Yes, a rapid temperature decrease causes the largest power increase when MTC is most negative, which is at EOL.

 

[Hiddencamper]

Yes, a rapid temperature decrease causes the largest power increase when MTC is most negative, which is at EOL.

Is this different between BWR/PWR?

In a BWR, the MTC becomes LESS negative (closer to positive, and at some points, completely positive) as you go from BOC to EOC. My BWR has a positive temperature coefficient about 1/2 way through the cycle.

 

[mudweez0009]

Yes, a rapid temperature decrease causes the largest power increase when MTC is most negative, which is at EOL.

So in a PWR, MTC is more positive at BOL, and more negative at EOL, meaning reactivity is highest at BOL than EOL, which makes sense to me.

I found that MTC = change in reactivity per degree change of coolant... Dont know how to use it to understand better though..

 

[QuantumPion]

Is this different between BWR/PWR?

In a BWR, the MTC becomes LESS negative (closer to positive, and at some points, completely positive) as you go from BOC to EOC. My BWR has a positive temperature coefficient about 1/2 way through the cycle.

In normal PWR land, the power is concentrated more towards the bottom early in cycle because that's where the coolant is coldest/densest. As the fuel burns up in the bottom, power shifts more towards the middle where the water is hotter and less dense, therefore more undermoderated and more negative MTC.

I have no idea what goes on in crazy BWR land with your axial zoning this and rodded that

 

[Hiddencamper]

Doing some more checkup.

MTC is the change in reactivity for a change in degree temperature.

I'm pretty sure it becomes less negative as you progress from beginning to end of cycle.

The SLOPE of the MTC curve also becomes larger towards EOC (Slope is what determines the overall reactivity effect due to a change in temperature).

See this image for what I'm talking about.

https://drive.google.com/file/d/0B-HOaUJyFtPhNkhaZVJIaGF3UmM/edit?usp=sharing

I guess I'm not completely understanding how this works in PWRs. A "More Negative" MTC would make me think that MSLB at EOC is LESS severe. Or is it MORE severe because there is more reactivity worth bound up in the MTC? I would think that your MTC would have LESS worth at end of cycle, because you are in a region with warmer/hotter water already...sorry lot of thoughts/questions

Sorry if I'm making this confusing.

 

[QuantumPion]

Doing some more checkup.

MTC is the change in reactivity for a change in degree temperature.

I'm pretty sure it becomes less negative as you progress from beginning to end of cycle.

The SLOPE of the MTC curve also becomes larger towards EOC (Slope is what determines the overall reactivity effect due to a change in temperature).

See this image for what I'm talking about.

https://drive.google.com/file/d/0B-HOaUJyFtPhNkhaZVJIaGF3UmM/edit?usp=sharing

I guess I'm not completely understanding how this works in PWRs. A "More Negative" MTC would make me think that MSLB at EOC is LESS severe. Or is it MORE severe because there is more reactivity worth bound up in the MTC?

Sorry if I'm making this confusing.

If you have a large negative MTC, then a small increase in power causes a large decrease in reactivity (good for normal operation). However, this also means a small decrease in coolant temperature causes a large increase in reactivity (bad for main steam line break - where you have a large, rapid decrease in temperature causing a very large increase in reactivity and power).

 

[Hiddencamper]

If you have a large negative MTC, then a small increase in power causes a large decrease in reactivity (good for normal operation). However, this also means a small decrease in coolant temperature causes a large increase in reactivity (bad for main steam line break - where you have a large, rapid decrease in temperature causing a very large increase in reactivity and power).

Ok, I think we are thinking the same thing.

More or less, the change in reactivity due to a change in moderator temperature at EOC will be more severe than at beginning of cycle.

With regards to the absolute magnitude/value of reactivity tied up in moderator temperature, I've been taught that the absolute value of reactivity due to temperature is less negative towards EOC. However that's really outside the scope of the question. The NRC fundamentals PWR question bank, used for the PWR licensed operator exam, says the same thing with regards to absolute value of reactivity.

Our crazy BWRs, depending on fuel design, can actually become overmoderated below 300 degrees F and have positive MTC. This happens when you go to long cycle + power uprate + new fuel designs.

Edit: Jim's post is great stuff. That makes a lot of sense.

 

[jim hardy]

PWR has competing temperature effects in moderator.
Water expands with temperature,separating the molecules which lessens its effectiveness at slowing down neutrons. That gives it a healthy negative temperature coefficient.
When boron is added to the water, the boron atoms also separate with temperature making them less apt to absorb neutrons. That effect is opposite , making the overall moderator temperature coefficient less negative at beginning of life when there's maximum boron..

Steam break cools the reactor
so the worry is that with a large negative temperature coefficient it'll return to power , though at lower than normal operating temperature , and continue feeding steam into containment if that's where the break is. That's why SI trips feeedwater pumps too.



From a Westinghouse AP1000 description:

The moderator temperature (density) coefficient is defined as the change in reactivity per degree change in the moderator temperature. Generally, the effects of the changes in moderator density and the temperature are considered together.
The soluble boron used in the reactor as a means of reactivity control also has an effect on the
moderator density coefficient, since the soluble boron density and the water density are decreased when the coolant temperature rises.
A decrease in the soluble boron density introduces a positive component in the moderator coefficient. If the concentration of soluble boron is large enough, the net value of the coefficient may be positive.
The initial core hot boron concentration is sufficiently low that the moderator temperature
coefficient is negative at operating temperatures with the burnable absorber loading specified.
Discrete or integral fuel burnable absorbers can be used in reload cores to confirm the moderator
temperature coefficient is negative over the range of power operation.
The effect of control rods is to make the moderator coefficient more negative, since the thermal neutron mean free path, and hence the volume affected by the control rods, increase with an increase in temperature.
With burnup, the moderator coefficient becomes more negative, primarily as a result of boric acid
dilution, but also to a significant extent from the effects of the buildup of plutonium and fission
products

http://www.nrc.gov/reactors/new-reactors/design-cert/ap1000/dcd/Tier 2/Chapter 4/4-3_r14.pdf

If you need some reading on the subject to put you to sleep, here's a study of steam breaks ...
https://www.oecd-nea.org/science/docs/2000/nsc-doc2000-21.pdf
I was going to quip "Rip Van Winkle must've written this", but after i got used to the awkward layout it became quite interesting...

One of the major concerns for the MSLB transient is the return to power and criticality in the latter half of the transient.
Because of this concern, the MSLB scenario is based on assumptions that conservatively maximise the consequences for a return to power. These assumptions, along with a detailed description of the reference problem, can be found in the following paragraphs...
...
...
The initial steady state problem places the reactor at 650 effective full power days (EFPD), end of cycle (EOC), with a boron concentration of 5 ppm, average core exposure of 24.58 GWD/MT, and equilibrium Xe and Sm concentrations. The initial operating conditions are as follows: the initial RCS pressure is 2 170 psia (the normal operating value), the initial pressuriser liquid level is set to 220 temperature compensated inches (a typical HFP level) and the initial cold leg temperature is at the normal value of 557 °F, which compensates for instrument error. Table 2.1 gives a detailed description of the initial steady state conditions for this transient...

page 53 of 136 has graphs of power vs time for various simulated steamline breaks. Looks like a return to ~ ten or twenty percent power after about a minute.

 

[mudweez0009]

I'm still a little confused. So from you guys' debates, I got that it is more severe at EOL. Maybe it comes down to me not having a full grasp on MTC.

My understanding of reactivity:

• I get that reactivity is based on Keff, and that if Keff>1 then the value of reactivity is positive, and if Keff <1 then the value of reactivity is negative.

• Reactivity is affected by temperature, among other things.

• The amount of reactivity change per degree change in moderator temperature is the moderator temperature coefficient.

• (From Jim's post): If moderator temp increases (which is just water flowing through the core?), then the water expands, causing it to not slow down neutrons, meaning more neutrons escape rather than being captured for further fission. So this why Jim said it gives a negative temp coefficient, because if less neutrons are being fissioned, then the reactivity value based on Keff decreases.

• I found a graph relating MTC to moderator temp, showing that MTC decreases as moderator temp increases. Below the graph it says that at moderator temp of room temperature, MTC is about -5 to -10 pcm/K and decreases down to about -25 pcm/K at operating temperature. Then at EOC, it can increase back up, even to being slightly positive.

I'm not sure from what I know how a reactivity decrease (from moderator temp increase) affects MTC any more at EOL than BOL, or vice versa.

I get that MTC is a function of reactivity, as it depends on reactivity change in the equation, but from that, can it be clarified what it means to discuss both reactivity AND MTC in terms of "positive" and "negative"? I feel it should be said in terms of "Less positive or more negative" OR "more positive or less negative"...

 

[QuantumPion]

I'm still a little confused. So from you guys' debates, I got that it is more severe at EOL. Maybe it comes down to me not having a full grasp on MTC.

My understanding of reactivity:

• I get that reactivity is based on Keff, and that if Keff>1 then the value of reactivity is positive, and if Keff <1 then the value of reactivity is negative.

• Reactivity is affected by temperature, among other things.

• The amount of reactivity change per degree change in moderator temperature is the moderator temperature coefficient.

• (From Jim's post): If moderator temp increases (which is just water flowing through the core?), then the water expands, causing it to not slow down neutrons, meaning more neutrons escape rather than being captured for further fission. So this why Jim said it gives a negative temp coefficient, because if less neutrons are being fissioned, then the reactivity value based on Keff decreases.

• I found a graph relating MTC to moderator temp, showing that MTC decreases as moderator temp increases. Below the graph it says that at moderator temp of room temperature, MTC is about -5 to -10 pcm/K and decreases down to about -25 pcm/K at operating temperature. Then at EOC, it can increase back up, even to being slightly positive.

I'm not sure from what I know how a reactivity decrease (from moderator temp increase) affects MTC any more at EOL than BOL, or vice versa.

I get that MTC is a function of reactivity, as it depends on reactivity change in the equation, but from that, can it be clarified what it means to discuss both reactivity AND MTC in terms of "positive" and "negative"? I feel it should be said in terms of "Less positive or more negative" OR "more positive or less negative"...

Reactivity is usually expressed in units of pcm (percent mille). So a reactivity of 10 pcm means k-effective is 1.00010 (i.e. the neutron flux increases by a factor of 1.00010 per neutron generation). MTC is expressed in units of pcm per degree F, e.g. -1 pcm/F would mean if moderator temperature decreases by 1 F than reactivity will increase by 1 pcm. Does this help?

Note that a change in reactivity does not affect MTC directly. Only changes in temperature or fuel composition do. Reactivity is simply the rate of change of the neutron population.

 

[jim hardy]

Then at EOC, it can increase back up, even to being slightly positive.

can you post link to that graph?

Starting at page 77 of this document (linked earlier) are graphs , and from a pretty credible source..
http://www.nrc.gov/reactors/new-reactors/design-cert/ap1000/dcd/Tier 2/Chapter 4/4-3_r14.pdf

Figure 4.3.24 on page 81 of 89 shows it going more negative as the cycle progresses (BOC is no burnup left side of chart, EOC is on right)
MWD/MTU is acronym for "Megawatt Days per Metric Ton of Uranium", a strange unit that's convenient for reactor engineer types.

I get that MTC is a function of reactivity, as it depends on reactivity change in the equation,but from that, can it be clarified what it means to discuss both reactivity AND MTC in terms of "positive" and "negative"?

ahhh here's why i was disappointed that teacher didn't do some reactivity calculations.

Keff is calculated as a product of probabilities.

Pages 9 to 16 of 65 here are a good rundown http://www.cedengineering.com/upload/reactor%20nuclear%20parameters.pdf


basically K = ε X p X f X η

where:
ε is a number very near 1.0 that represents probability of a fission being thermal not fast(fast fission factor)
p is a number near .95 to 1.0 that represents probability of a neutron not getting absorbed in a non-fission capture in the fuel (resonance escape probability) ie probability that if it gets absorbed by fuel it'll cause a fission
f is a number somewhat less than 1 representing probability that a thermal neutron is absorbed in the fuel instead of something else like a control rod(thermal utilization factor),
η is a number representing the number of neutrons produced per fission, and runs a little more than 2. It varies from ~2 to 2.5 depending on fuel isotopes.

When the result of that four term product is exactly 1, you are exactly critical.
So i guess i told a white lie up above - it's a product of three probabilities and a constant for isotoope.

Poison like boron or rods affects f
ε and p change with burnup
moderator temperature affects how far a neutron must travel to get slowed down, hence it affects f and probably p (perhaps a genuine nuclear type can help me here?)
AND temperature affects how much boron is in the core which is definitely f.

A lot to absorb i know

but your curiosity is invigorating to an old guy like me who loves this stuff. Secondhand Lion, you know...

Keep up the good work !

 

[mudweez0009]

Okay, thanks guys.

can you post link to that graph?

http://www.energy.kth.se/courses/4A1627/Material2005/PhysicsPart/02%20Reactivity%20Rev%200.pdf

Pages 27 and 28

 

[Astronuc]

Okay, thanks guys.



http://www.energy.kth.se/courses/4A1627/Material2005/PhysicsPart/02%20Reactivity%20Rev%200.pdf

Pages 27 and 28

The example is for a BWR rather than a PWR. The curve shows MTC as a function of moderator (coolant) temperature, and in a BWR, the coolant does not use soluble boron as is the case in a PWR. Also, BWRs operate at lower coolant temperature (Tsat ~ 286) and at shutdown, the temperature decreases to ~100 C or less when the head is removed. However, at shutdown in a BWR, all the control rods are in the core at shutdown.

For a PWR, the soluble boron is mostly gone, and one can use feedwater temperature reduction to add some reactivity to the core to extend the cycle during coastdown. Also, at EOC, there is a fair amount of Pu-239/-241 built up in high burnup fuel, and the burnable poison in the fresh fuel (feed batch, which becomes once-burned at EOC) is depleted.

MTC will also depend on the enrichment and use of burnable poison, which are somewhat affected by core design, e.g., batch size and cycle length/exposure. I've seem some plants on 24 month cycles have issues with + MTC, however, I can't remember if it was BOC, MOC or EOC, and it might have been during a transition cycle.

The pdf that Jim cites states, "if a reactor is over-moderated, it will have a positive MTC." It could, but not necessarily so. IIRC, use of enriched (in B-10) boron is one method of reducing MTC.

 

[jim hardy]

Thanks Astro for the insight.

MTC will also depend on the enrichment and use of burnable poison, which are somewhat affected by core design, e.g., batch size and cycle length/exposure.

So the designer can tweak core design to somewhat bend MTC where he wants it... makes perfect sense..
I could accept that accumulation of Pu as cycle progresses might change η, and depletion of U235 might affect ε
but that's something i'd ask a reactor engineer about rather than pretend i know .

thanks again.

 

[mudweez0009]

Okay this is going to be another random question, but this is evidently the trend of this course..
I actually created a new thread for this topic for my specific questions, but it doesn't seem to be attracting anyone. I figured while I have a few people's attention, I may as well post it here, even though it is off our current topic.
I will admit upfront that these are directly from my homework. From this thread, I hope I have proven I am genuinely trying to learn the material and not just ask for answers. But I need help, and can't find any equations anywhere (yes, we got calc problems.. and yes, weren't given reference material).

The thread link: (https://www.physicsforums.com/showthread.php?t=747913#taglist)

You can go there if you choose, or we can continue it here, as this conversation has definitely evolved throughout its life anyway. If you choose to stay here, here is my post in the linked thread..

I am having issues with some problems relating to a plant theoretically shut down on natural circulation, and calculating the core ΔT and natural circulation flow rate.. Can anyone provide some equations or theory I could use to assist me? I'm not familiar with this material and have spent hours searching Google and cannot find much.

Problem 1:
Givens:
Initial Decay heat = 2.5% rated thermal power.
Initial Core ΔT = 14 deg F
Final Decay Heat = 1% rated thermal power
Find: Final Core ΔT.

Problem 2:
Givens:
Initial natural circulation flow rate = 3.5% full power flow rate.
Initial Core ΔT = 15 deg F
Final Core ΔT = 8 deg F
Find: Final natural circulation flow rate

 

[QuantumPion]

Homework questions should be posted in the homework help forum, they have specific guidelines about showing your work and such.

That being said, I'd need more information about what topics you've been covering in this course specifically in order to know where to start.

 

[mudweez0009]

Homework questions should be posted in the homework help forum, they have specific guidelines about showing your work and such.

That being said, I'd need more information about what topics you've been covering in this course specifically in order to know where to start.

I made my account right when I started this thread, so I'm unfamiliar of the rules. Thanks for the heads up.

That said, we literally are covering big picture theory on how the plant systems are integrated.. The course is taught once a week. We spent a day for each the steam generator, turbine, CVCS system, electric stuff (totally lost me), cooling water/condensor, etc. For this specific class, we were talking about natural circulation. We then get assigned very specific homework such as this, and most of us, especially us non-nuclear engineers, don't know how to do it. I would love to show you the slides from the course to prove how useless they are.

--------
edit: The template for homework requires "useful equations" and "an attempt at solving the problem".. Kind of hard to do when I don't know any equations.

 

[jim hardy]

wow that's sketchy.

It's a mechanical engineering problem, not a nuclear one.
It assumes a-priori knowledge about the system.my approach would be:
(and a genuine mechanical engineer might well have a better one)
in natural circulation the driving force for flow is the difference in density between:
water in the hotleg going up to the steam generator
and water in coldleg returning from it.

And teacher didnt give that height.

so driving head is Δ \rho g h

with height unknown.

%power divided by ΔT should be pretty linear with flow, it's just moving heat,and flow should be in proportion to √driving head (bernoulli)
is density of subcooled water fairly linear with temperature over the small range of ΔT ? (Teacher didn't give us pressure...)
If so then ratio flow/√ΔT = constant should be reasonable approximation.Bounce that around the ME circle in your class and see if anythiing comes of it.
Maybe try those equations in homework thread too.

In range of .01 to .1, just how nonlinear is √ function? Not very, as i recall
i'd run trial calcs on both flow/ΔT and flow/√ΔT ...

 

[QuantumPion]

Yeah my confusion is similar. Natural circulation is a complicated process dependent on many variables. My thoughts of approaching it was based on a simple energy balance. But they don't tell you the temperature so there is insufficient information to determine anything. My only guess is that it is a super oversimplified proportion problem.